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Answer 1 · 2017-01-18T20:09:28

The primary kinetic isotope effect does not affect the reaction mechanism; rather, it gives information on the reaction mechanism.

Kinetic Isotope Effect

The kinetic isotope effect (KIE) refers to the change in rate caused by an isotopic substation in a molecule.

Frequently, the KIE refers to the effect of substituting $"D"$ for $"H"$ (the $k_"H"//k_"D"$ ratio).

The $"C-H"$ and $"C-D"$ vibrations are quantized per the condition

$E = (n + 1/2)hf$ , where $n = 0, 1, 2, …$

At ordinary temperatures, most of the molecules are in the $n = 0$ state.

We see the maximum KIE when a $"C-H"$ bond is being broken in the transition state, because the vibrational mode disappears.

Since $E_0("H") > E_0("D")$ , $E_"a""(H") < E_"a""(D)"$ and $k_"H" > k_"d"$ .

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This is called a primary kinetic isotope effect.

The calculated value of $k_"H" // k_"d" ≈ 7$ .

A secondary KIE occurs when the isotope is substituted at a position next to the bond being broken.

Usually, for a secondary KIE, $k_"H"//k_"D" < 1.5$ .

Mechanistic information from KIEs

KIEs give useful information about the rate determining step in a mechanism.

Case 1

$"CH"_3"CH"_2"CH"_2"Br" stackrelcolor(blue)( "EtO"^"-", "EtOH"color(white)(m)) (→) "CH"_3"CH=CH"_2$

$"CH"_3"CD"_2"CH"_2"Br" stackrelcolor(blue)( "EtO"^"-", "EtOH"color(white)(m)) (→) "CH"_3"CD=CH"_2$

$k_"H"//k_"D" = 6.7$

This is consistent with an $"E2"$ elimination in which the $"C-H/D"$ bond is being broken in the rate determining step.

Case 2

$"CH"_3"CH"_2"C"("CH"_3)_2"Br" stackrelcolor(blue)( "H"_2"O", Δcolor(white)(m)) (→) "CH"_3"CH=C"("CH"_3)_2$

$"CH"_3"CD"_2"C"("CH"_3)_2"Br" stackrelcolor(blue)( "H"_2"O", Δcolor(white)(m)) (→) "CH"_3"CD= C"("CH"_3)_2$

$k_"H"//k_"D" = 1.4$

This is consistent with an $"E1"$ elimination in which the $"C-H(D)"$ bond is not being broken in the rate determining step.